# How does this calculation of $\int_0^\infty(\sin^2x)/x^2\ dx$ work?

I'm having trouble with two steps in a calculation of

$$\int_0^\infty\left(\frac{\sin x}{x}\right)^2\ dx$$

in a book.

• They take the contours $C_R$ composed of upper half-circles $H_r=\{re^{i\theta}\,:\,\theta\in[0,\pi]\}$ and $G_R=\{Re^{i\theta}\,:\,\theta\in[0,\pi]\},$ where $r<R,$ and the real intervals $I_{r,R}=[-R,-r]$ and $J_{r,R}=[r,R].$

• They notice that $2\sin^2x=\Re(1-e^{2ix})$ for $x\in\Bbb R.$ This allows them to consider the function $$f(z)=\frac{1-e^{2iz}}{z^2}.$$

• Now they do something I don't understand. They say that $$\int_{I_{r,R}}f(z)\ dz=\int_{J_{r,R}}f(z)\ dz.$$ How is this true? I have $$\int_{I_{r,R}}f(z)\ dz=\int_{-R}^{-r}\frac{1-e^{2ix}}{x^2}\ dx=-\int_{R}^{r}\frac{1-e^{-2ix}}{x^2}\ dx=\int_{r}^{R}\frac{1-e^{-2ix}}{x^2}\ dx,$$ but $$\int_{J_{r,R}}f(z)\ dz=\int_{r}^{R}\frac{1-e^{2ix}}{x^2}\ dx$$ without the minus in the exponent... And the exponential function isn't even.

• Now they say that $$|\int_{G_R}f(z)|\leq\frac{2\pi}R\to 0$$ as $R\to\infty.$ I've been looking at it for over an hour, and I can't see this. Maybe I won't post my calculations because the post is already long. It looks completely false to me. Shouldn't the exponential function in the numerator dominate the denominator? I tried to use the triangle inequality and the inequality $|e^z|\leq e^{|z|}$ after substituting $z=Re^{i\theta}.$

• Now it's easy to finish the calculation, but since I don't understand the previous steps, it doesn't count.

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The two integrals (over $I_{r,R}$ and $J_{r,R}$) don't look equal to me, but their real parts will be equal, since the real part of $e^{2iz}$ is even. That is probably all you need (I haven't checked in detail, but remember you are starting with real integral).
In the upper half plane $e^{2iz}=e^{2ix}\cdot e^{-2y}$. The first factor has modulus $1$, the second is less than $1$. I think that deals with your second problem.